Solution

I. 2y² - 3y – 14 = 0 2y² - 7y  + 4 y – 14 = 0 y (2 y – 7) + 2( 2 y – 7) = 0 (y + 2) (2 y – 7) = 0 ∴ y = -2, 7/2 II. 3x² - 7x + 4 = 0 3x² - 4x - 3 x  + 4 = 0 x (3 x – 4) – 1 (3 x – 4) = 0 ∴ x = 1, 4/3 Hence, relationship cannot be established between x and y Alternate Method: if signs of quadratic equation is - ve and -ve respectively then the roots of equation will be -ve and +ve. (note: -ve sign will come in smaller root) So, roots of first equation = y = -2, 7/2 if signs of quadratic equation is - ve and +ve respectively then the roots of equation will be +ve and +ve. So, roots of second equation = x = 1, 4/3 After comparing we can conclude that relationship cannot be established.